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The satellite will return to the planet surface when the graviataional pull on it is GREATER than the required centripetal force to keep it in orbit. That is,
G(m1)(m2)/(r2)^2 > (m2)(Vb)^2/r2 where G is the Universal Gravitational Constant
[Note: when the left term equals to the right term, the satelllite will stay in orbit. If the left...
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最佳解答：
The satellite will return to the planet surface when the graviataional pull on it is GREATER than the required centripetal force to keep it in orbit. That is,
G(m1)(m2)/(r2)^2 > (m2)(Vb)^2/r2 where G is the Universal Gravitational Constant
[Note: when the left term equals to the right term, the satelllite will stay in orbit. If the left term is smaller than the right term, the satellite will fly away from point B]
thus, (Vb)^2 < G(m1)/r2
or (Vb)^2 < G(m1)/X.r1 ------------------ (1)
From conservation of mechanical energy,
loss of kinetic energy of satellite = gain in potential energy of satellite
i.e. (1/2)(m2)(Va)^2 - (1/2)(m2)(Vb)^2 = G(m1)(m2)(1/r1 - 1/r2)
Because r2 = X(r1), we have, after simplification,
(Va)^2 - (Vb)^2 = 2G(m1)(X - 1)/(X.r1)
i.e. G(m1)/(X.r1) = [(Va)^2 - (Vb)^2]/(2.(X - 1))
Substitute into (1):
(Vb)^2 < [(Va)^2 - (Vb)^2]/(2.(X - 1))
Divide both sides by (Vb)^2,
1 < [(Va/Vb)^2 -1]/(2.(X -1))
Inverting the fraction on the right-hand-side, then
2(X -1)/[(Va/Vb)^2 - 1] < 1 --------------- (2)
I suppose you have already found the expression for Vb/Va in part (d). Let this be f, say.
i.e. Vb/Va = f where f is a function of "alpha" and X
Hence, inequality (2) becomes,
2(X -1)/[(1/f)^2 - 1] < 1
The expression on the left-hand-side involves X and "alpha" only. I just leave it for you to complete the derivation, as it involves only algebric manipulation.
Hope the above would help.

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In your equation, I suppose the parameter θ is the angle between the rod and the top of the car. In so doing, your equation is actually describing an oscillation about the top of the car (because θ is measured from the top of the car).
But the rod is, in fact, oscillating about its equilibrium position, which is at an angle θ with the top of...
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In your equation, I suppose the parameter θ is the angle between the rod and the top of the car. In so doing, your equation is actually describing an oscillation about the top of the car (because θ is measured from the top of the car).
But the rod is, in fact, oscillating about its equilibrium position, which is at an angle θ with the top of the car.
Moreover, according to your equation, the force mg.sin(θ) and the force mA.cos(θ) are acting along the length of the rod, but in opposite directions. Since the two forces pass through the pivot, there should be no torque produced. But you have made a wrong assumption that the two forces are acting perpendicularly to the rod.

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From your given link: https://en.wikipedia.org/wiki/Optical_medium , an "optical medium" is a substance through which electromagnetic waves propagate.
Following this definition, metals (including alloys) are not optical media, as metals are conductors and which electromagnetic waves cannot pass through.
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Reply...
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最佳解答：
From your given link: https://en.wikipedia.org/wiki/Optical_medium , an "optical medium" is a substance through which electromagnetic waves propagate.
Following this definition, metals (including alloys) are not optical media, as metals are conductors and which electromagnetic waves cannot pass through.
----------------------
Reply to your question in "意見":
You are asking, in your question, the material substances that are not optical medium. Copper, clearly, is NOT an optical medium, as electromagnetic waves cannot pass through (see the definition of optical medium in your second link).
The reflection of light on polished copper surface indicates that light wave cannot pass through the substance.

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To find the Norton current (In), assume pionts a and b are connected together.
Hence, current through the 9-ohm resistor = 18/9 A = 2 A
Becasue the 2-ohm resistor is in series with the 3-ohm and 6-ohm resistors in parallel (after a and b are connected), the equivalent resistance (Req) of this branch is,
Req = (2 + 3//6) ohms (here the...
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最佳解答：
To find the Norton current (In), assume pionts a and b are connected together.
Hence, current through the 9-ohm resistor = 18/9 A = 2 A
Becasue the 2-ohm resistor is in series with the 3-ohm and 6-ohm resistors in parallel (after a and b are connected), the equivalent resistance (Req) of this branch is,
Req = (2 + 3//6) ohms (here the symbol 3//6 means "3-ohm resistor connected in parallel with 6-ohm resistor).
Thus, Req = [2 + 3x6/(3+6)] ohms = (2 + 2) ohms = 4 ohms
Hence, voltage across resistance 3//6 (which equals to 2 ohms) = 18 x [2/(2+2)] v = 9 v
or voltage across 3-ohm resistor = 9 v (because 3-ohm and 6-ohm are in parallel)
Current through 3-ohm resistor = 9/3 A = 3 A
Since the current flows through a and b is the sum of currents from the 9-ohm resistor and 3-ohm resistor, we have,
the Norton current In = (2 + 3) A = 5 A
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To find the Norton resistance (Rn), assume the 18v source is in short-circuit. Here, we have two main branches of resistance. The 1st branch is simply the 9-ohm resistor. The 2nd branch is the 3-ohm resistor in series with the 2-ohm and 6-ohm resistors in parale.
First, calculate the resistance of the 2nd branch R2.
R2 = (3 + 2//6) ohms = (3 + 2x6/(2+6)) ohms = (3 + 1.5) ohms = 4.5 ohms
Therefore, the Norton equivalent resistance Rn = (9//4.5) ohms = 9x4.5/(9 + 4.5) ohms
= 3 ohms

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1. No, water molecules below the surface do move. Water molecules in a "surface wave" move in circular orbits. The diameter of the orbit decreases with distance below the water surface. At a verical distance of one wavelength below the surface, the orbit becomes nearly zero, i.e. water molecules stop to move at such depth and...
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1. No, water molecules below the surface do move. Water molecules in a "surface wave" move in circular orbits. The diameter of the orbit decreases with distance below the water surface. At a verical distance of one wavelength below the surface, the orbit becomes nearly zero, i.e. water molecules stop to move at such depth and below.
2. As said above, water molecules move in circular orbits in a water wave. A circular orbit is indeed a transverse wave and a longitudinal wave superimposed on each other. Hence, water molecules have both vertical and horizontal displacements. Note that these two displacements are 90 degrees (or a quarter of wavelength) out of phase.
Because the logitudinal (i.e. horizontal) motion of molecules and the transverse (i.e. vertical) motion of molecules are 90 degrees out of phase. At the wave crest and trough, the vertical displacement is maximum, the horinztonal displace is zero. Hence, at the mid-point between the crest and trough (i.e. the y-displacement, which indicates the vertical displacement, of the wave is zero), the horizontal displacement, which has a phase difference of 90 degrees with the vertical displacement, will be at maximum.
Perhaps we can look it on another way using Simple Harmonic Motion (SHM) of water molecules. Just consider the longitudinal motion of the water molecules. These molecules perform Simple Harmonic Motion, the the equilbrium positions are located at the trough and crest of the water wave. Similarly, for the transverse motion of molecules, but the equilibrium positions are at the mid-point between the crest and trough.
Because the slope on the shape of the water wave indicates the speed of the water molecules on the transeverse (vertical) motion of molcules, the slope is greatest at the mid-point between the crest and trough. Thus, the vertical speed of molecules is maximum at this point. The horizotnal speed of molecules is thus minimum (i.e. zero) at such point, because the two motions are 90 degrees out of phase. From theroy in Simple Harmonic Motion, a zero velocity implies that the displacement is masximum.
The following web-page gives a very clear explanation on water waves. You may take a look on it:
http://labman.phys.utk.edu/phys221core/modules/m12/Water_waves.html

(1) Using Boyle's Law,
30000 x 8 = P x (8 + 4.3)
where P is the final pressure after the two steel vessels are connected.
Hence, P = 30000 x 8/(8 + 4.3) psi = 19,512 psi
(2) Using the Ideal Gas Equation, assume the temperature of the air inside the vessel T remains unchanged, the no. of moles of air in each vessel, before they are connected, ...
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(1) Using Boyle's Law,
30000 x 8 = P x (8 + 4.3)
where P is the final pressure after the two steel vessels are connected.
Hence, P = 30000 x 8/(8 + 4.3) psi = 19,512 psi
(2) Using the Ideal Gas Equation, assume the temperature of the air inside the vessel T remains unchanged, the no. of moles of air in each vessel, before they are connected, is given by,
For the 8L vessel, no. of moles of air inside n1 = where R is the Universal Gas Constant.8 x 30000/RT
For the 4.3L vessel, no. of moles of air inside n2 = 4.3 x 10000/RT
Hence, total mass (no. of moles) of air in the two vessles before connected = n1 + n2 = [(8 x 30000) + (4.3 x 10000)]/RT
Mass (no. of moles n) of air after the vessels are connected
n = (Pf)(8 + 4.3)/RT
where Pf is the final pressure after connection.
Using Law of Conservation of Mass, no. of moles of air in the two vessels before and after connection are the same.
i.e. n1 + n2 = n
[(8 x 30000) + (4.3 x 10000)]/RT = (Pf)(8 + 4.3)/RT
solve for Pf gives the final pressure = 23,008 psi.

(a) Let Q be the amount of heat conducted through unit cross-sectional area of the wall in unit time.
Then, Q = -k(dT/dx) -------------- (1)
where k is the thermal conductivity of the wall (= 0.77 W/mK)
T is the temperature at distance x in the wall.
x is the distance from the inside surface of the wall.
[The -ve sign indicates that heat flows...
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Sports Mechanics is not my specialty, but I would like to give my view on this problem.
Stance I : Initial Impact Peak
This happens when the incoming ball is first caught by the racket .
Phase II: Secondary Impact Peak
When the incoming ball first collides with the racket, it will rebound. The player then swings the racket to catch up with the...
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Sports Mechanics is not my specialty, but I would like to give my view on this problem.
Stance I : Initial Impact Peak
This happens when the incoming ball is first caught by the racket .
Phase II: Secondary Impact Peak
When the incoming ball first collides with the racket, it will rebound. The player then swings the racket to catch up with the ball again. This is the second impact.
Phase III: Weight Acceptance
As far as the player swings the racket, the ball is being pushed forward rapidly, causing it to accelerate At such moment, the force mainly comes from the swinging arm of the player. Hence, the curve on vGRF (vertical ground reaction force) varies not much.
Phase IV: Drive-off
Here, the racket provides a final strike onto the ball and makes it flying away from the racket .

If the force is applied in direction as given in your diagram, the force component pushing the wedge to move horizontally will be zero.
But I wonder if you have misinterpret the question. It says that the force is "...垂直的壓在斜面上..", it may mean the force is perpendicular to the slant surface of the wedge. In that case, the component of...
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If the force is applied in direction as given in your diagram, the force component pushing the wedge to move horizontally will be zero.
But I wonder if you have misinterpret the question. It says that the force is "...垂直的壓在斜面上..", it may mean the force is perpendicular to the slant surface of the wedge. In that case, the component of force pushing the wedge to the right is 100.sin(50) N = 76.6 N

The Right Hand Grip rule serves TWO purposes. One is to find the direction of magnetic field surrounding a current in a wire. The other is to find the "direction" of the North pole in a current carrying coil.
I suppose you ask for the first purpose of using the rule. It says that if the thumb describes the direction of current, then the...
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The Right Hand Grip rule serves TWO purposes. One is to find the direction of magnetic field surrounding a current in a wire. The other is to find the "direction" of the North pole in a current carrying coil.
I suppose you ask for the first purpose of using the rule. It says that if the thumb describes the direction of current, then the other fingers describe the direction of the magnetic field.
It means that if you are loonking along the direction of the electric current at the end of the wire, the mangetic field will be in the "clockwise direction". On the contrary, if you are looking at the direction against the direction of current flow, then the direction of magnetic field will be in the "anticlockwise direction".

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You are welcome !
It's my pleasure thast I could share my knowledge and experience in physics with you, and with other candidates in this forum. Your success is attributed to your hard work and willingness to learn.
Congratulation to your getting a high score in physics. Wish you would enjoy studying physics at university.
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最佳解答：
You are welcome !
It's my pleasure thast I could share my knowledge and experience in physics with you, and with other candidates in this forum. Your success is attributed to your hard work and willingness to learn.
Congratulation to your getting a high score in physics. Wish you would enjoy studying physics at university.

The ignition temperature (also refers as auto-ignition temperature) for paper is commonly taken as 233 degrees Centigrade (or 451' F). Since there are different kinds of paper with various material composition, volume, density and shape, the ignition temperature may varies from around 218'C to 246'C.
Reference:...
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The ignition temperature (also refers as auto-ignition temperature) for paper is commonly taken as 233 degrees Centigrade (or 451' F). Since there are different kinds of paper with various material composition, volume, density and shape, the ignition temperature may varies from around 218'C to 246'C.
Reference: http://hypertextbook.com/facts/2003/LewisChung.shtml
It takes a few minutes for a sheet of paper to burst into flames when it is put at its ignition temperature, but it would take much longer than that for a thick book. The dense material in the center of a book shunts heat away from the outside edges, preventing them from reaching the auto-ignition temperature.

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4. The equations involved in calculating the thrust on an inclined plane are:
Thrust on inclined plane Fr = pg(hc).A
where p is the density of water (= 1000 kg/m^3), g is the acceleration due to gravity (= 9.81 m/s^2), hc is the vertical depth, from water surface, of the centre of mass of the plane, A is the area of the plane.
For a...
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最佳解答：
4. The equations involved in calculating the thrust on an inclined plane are:
Thrust on inclined plane Fr = pg(hc).A
where p is the density of water (= 1000 kg/m^3), g is the acceleration due to gravity (= 9.81 m/s^2), hc is the vertical depth, from water surface, of the centre of mass of the plane, A is the area of the plane.
For a rectangular plane of height H and width W, the thrust acts at a distance (yr), measured in direction along the inclined plane from water surface, is
yr = yc + WH^3/(12.WH.yc)
where yc is the distance, measured in direction along the inclined plane from water surface, of the centre of mass of the plane, and
yc = hc/sin(a) where hc is the vertical depth from water surface of the centre of mass of the plane, and a is the inclined angle of the plane.
First consider the thrust acting on the left hand side of the gate,
Fr = 1000 x 9.81 x [4 - 1 x sin(60)] x (2 x 1) m = 61,489 N
The point of action of Fr is given by,
yr = yc + (1 x 2^3)/[12 x (1 x 2)yc]
where yc = [4 - 1 x sin(60)]/sin(60) m = 3.619 m
Hence, yr = 3.619 + (1 x 2^3)/[12 x (1 x 2) x 3.619] m = 3.711 m
Distance along the plane of Fr from the hinge A = yr - [(4 - 2.sin(60))/sin(60)] m
= 3.711 - [(4 - 2.sin(60))/sin(60)] m = 1.092 m
There the thrust on the left hand side of the gate is 61,489 N and acts at a distance of 1.092 m from the hinge at A.
Using the same method, the thrust on the right hand side of the gate Fr' can be found equals to 7,358 N and acts at a distance of 1.423 m from A.
Take moment about hinge A,
W.cos(60) x 1 + (Fr') x 1.423 = (Fr) x 1.092
i.e. W = (61489 x 1.092 - 7358 x 1.423)/cos(60) N = 113,351 N
5. You could use the same equations given above to solve this problem. Just try it yourself.